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AIR COMPRESSORS
11.1 Introduction
Air compressors are used for supplying high-pressure air. There are many uses of high-pressure air in the industry. The main uses of high-pressure (compressed) air are :
.. to drive compressed air engines (air motors) used in coal mines, .. to inject or spray fuel into the cylinder of a Diesel engine (air injection Diesel engine), to operate drills, hammers, air brakes for locomotives and railway carnages, water pumps and paint sprays, .. to start large (heavy) Diesel engines, .. to clean workshop machines, generators, automobile vehicles, etc., .. to operate blast furnaces, gas turbine plants, Bessemer convertors used in steel plants, etc., .. to cool large buildings and air crafts, and .. to supercharge I.C. engines. There are mainly two types of air compressors viz. reciprocating air compressors and rotary air compressors. Reciprocating air compressors are similar to reciprocating engines where a piston reciprocates inside a cylinder. In rotary air compressors, air is compressed due to rotation of impeller or blades inside a casing. Air compressors are driven by engines or electric motors. In this chapter the theory of reciprocating air compressors is discussed in details and principles of working of reciprocating compressed air motors and rotary compressors are explained in brief. 11.2 Reciprocating Air Compressors The principal parts of a reciprocating air compressor are the same as that for a engine. The reciprocating air compressor may be single-acting (air is admitted to one side of the piston only) or double-acting (air is admitted to each side of the piston alternatively), and may be single-stage or multi-stage. In a multi-stage compressor, the air is compressed in several stages instead of compressing the air fully in a single cylinder. This is equivalent to a number of compressors arranged in series. The pressure of air is increased in each stage. Single-stage compressors are used for delivery pressures upto 10 bar, three-stage compressors for pressure upto 200 bar and two-stage compressors for pressures in between 10 to 200 bar. The average piston speed of a reciprocating air compressor is limited to about 300 to 400 metres per minute to reduce friction wear. 11.3 Single-stage Air Compressor The sectional view of an air cooled, single-stage, single-acting reciprocating air compressor is shown in fig. 11-1. Both intake (suction) and discharge (delivery) valves
AIR COMPRESSORS 343
CyliMtr Piston
Fig. 11-1. Sectional view of a single- acting air cooled, reciprocating air compressor.
are disc type and are automatic in their action. They are opened and closed by difference in the air pressure acting on their two sides. When the pressures are equal on their two sides, they are kept closed by light springs. During the outward or suction stroke, the pressure in the cylinder falls below the atmospheric pressure as a result of which the intake valve opens and air is drawn from the atmosphere into the cylinder. During the inward or compression stroke, as a result of the piston action the pressure of the air in the cylinder gradually increases and reaches a value sufficiently above the receiver pressure. The high pressure of air, thus produced, overcomes the resistance of the spring on the discharge valve and causes the valve to open and discharge takes place from the cylinder to the receiver. The receiver is a simple vessel which acts as a storage tank. The compressor is driven by some form of prime mover (electric motor or engine). When the compressor is to be started against tank (receiver) pressure, the prime mover will have to supply very high starting torque. To avoid this, hand unloader (fig. 11-1) is used for releasing pressure from the compressor cylinder when the com- pressor is stopped.
11.3.1Indicator Diagram : The events described above can be conveniently represented by p -v diagram shown in fig. 11-2. The diagram is drawn for a compressor without clearance. During the suction stroke the charge of air is drawn into the cylinder along line 4-1 at constant pressure p i, which is slightly below that of the atmosphere. At point 1, the piston completes the suction (outward) stroke and starts on its return (compression) stroke. All valves being closed, the air is now compressed along the compression curve 1-2. At point 2, pressure P 2 is reached which is slightly higher than the pressure in the receiver. The discharge valve at this point opens and the delivery of the compressed air takes place along
line 2-3 at pressure p 2. The piston has now reached
the left hand end of the cylinder and again starts on its suction stroke and the pressure in the cylinder will be lowered again to p i and the cycle of operations will be repeated. The net work required for compression and Fig. 11-2. I heorotical indicator diagram of .delivery^ of the air per cycle is represented by the area i i O V a single-stage air compressor. l-Z -o -4 (tig. 1 1 -2 ).
The amount of work done on the air will depend upon the nature of the compression curve. If the compression occurs very rapidly in a non-conducting cylinder so that there is no heat transfer, the compression will be practically isentropic. If it is carried out slowly so that the heat of the compression is extracted from the air by the jacket cooling water, the compression will approach isothermal. However, in actual practice
AIR COMPRESSORS 345
= 0P 2V2 - PzV2 + P 2 V 2 - P M - " P M +P1^
n - n
^ ~ n _'-\ (P2V 2 - pivi) Joule per cycle
o r w = mR(T 2 - 7i) Joule per cycle
From eqn. (11.1a) taking pjy* outside the bracket,
.. (1 i:ia )
.. (11.1b)
Work required, W = n n - 1 P ivi^
1 _pw _
But for polytropic compression, p i v i n = p 2 V 2 n. Hence, — =
_v_
Substituting the value of £ in the above equation,
Work
w S ' s
v- required per cycle (or per revolution, if compressor is single-acting),
1 us W (^) = - - — n p i vt (^) X ' k ' 1 n
- 1 W [^ p^ i j
n — 1
P1V
c 1
( p z )
-----------
1
1
c
&
Joule per cycle .. (11.2) This equation gives the work required in Joules per cycle (or per revolution, if the compressor is single-acting) in compressing and delivering the air. It should be noted that units of pressure and volume in eqn. (11.2) are N/m2 or Pa and m3 respectively. .Indicated ^ (^) power of the compressor = — - - — W x N (^) J/sec. or W... ...(11.3)
where W = work required in Joules per cycle, and N - No. of cycles performed per minute - r.p.m. for single-acting compressor, if p iv i in eqn. (11.2) is substituted by mRTi, then work required per cycle,
n - (P z 1
1
c
( p i j
W = — mRTi n - 1
Work required per kg of air,
Joule per cycle and
W = n n - 1
RT,
' P i 1
1
c
[ p i j
Joule (^) ...(11.4)
Indicated power of the compressor = W x mass of air delivered per second J/sec. or W ... (11.5) where W = work required in Joules per kg of air.
(b) If the compression is isentropic (pvT = C), the index n will be replaced by y in eqn. (11.2) and eqn. (11.4), and then
346 ELEMENTS OF HEAT ENGINES Vol.l
Work required, W =
Work required per kg of air,
W =
Y - 1
— _p\v_ y -1 r
pst (^) 7 -1 (^) Joule per cycle ... (11.6a)
n
Y -
VPi J
Joule (11.6b)
(c) If the compression is isothermal (pv = C), then work required per cycle, W » area 1-2-3-4 (fig. 11-4) = area a-2-1-b plus area 0-a-2-3 minus area b-1-4-
Y
V 2.
W = P 1 V 1 loge
= P1V1 loge
+ P2V2 - P 1 V 1 rn P1V1 loge
py
Joule per cycle
(since P 2 V 2 - pivi)
... (11.7a)
Work required per kg of air, W - RTi loge v pi
Joule
(11.7b)
11.3.3 Approximation of Isothermal Compression : Although isothermal compression is economical, it is not possible to achieve it in practice. To have an isothermal compression, the compressor will have to be run extremely slow, while in practice it is driven at high speed so that as much air as possible in compressed in a given time. Since, there is saving of work by compressing air isothermally, it is necessary to make an attempt to obtain approximately an isothermal compression. Three methods are adopted to achieve this object while still running the compressor at high speed. The three methods adopted are : — cooling the air during compression by spraying cold water into the cylinder, — cooling the air during compression by circulating cold water through the cylinder jacket, and — adopting multi-stage compression with inter-stage cooling. Cold water spray : In this method, cold water is sprayed into the cylinder during compression. The cooling, thus done, reduces the temperature of the air and the compression curve will be approximately of the form pv1 = C. This means that the compression is brought nearer to isothermal which results in the saving of work. Water jacket : In this method, the heat of compression is extracted by circulating cold water in the cylinder jacket thereby keeping the temperature rise as small as possible. This keeps the compression near to isothermal as shown in fig. 11-3.
348 ELEMENTS OF HEAT ENGINES Vol.l
perfect-intercooling. The low pressure cylinder diagram is shown as p r-1 -2 ’-p2, and high pressure cylinder diagram as p2-2—3-p3. The reduction of work required due to intercooling is shown by the shaded are 2-3-3’-2’ (fig. 11-6). When cooling is perfect,
i.e., when air is cooled to intake temperature in the intercooler (T i = T 2 ), the point 2
will lie on the isothermal line 1 - 3" as shown in fig. 11-6. It may be noted that each stage will increase the pressure of air while the intake temperature Ti (corresponding to point 1) is maintained same at the end. The isothermal line during the process has been approximated as shown by the diagram, and the shaded area 2-3-3’-2’ shows the saving of work as a result of this approximated isothermal. 11.4.1 Imperfect-lntercoollng
Fig. 11-7. Combined indicator diagram of two-stage compression with imperfect intercooling.
: Figure 11-7 represents the indicator diagram of a two-stage air compression with imperfect-intercooling. Let the compression follow the law p i/1 = constant and the intercooling be incomplete (imperfect) so that the point 2 has not reached the isothermal line, i.e., point 2 does not lie on the isothermal curve 1-3”. Let p i in N/m2 = Pa and vi in m3 represent
condition of air entering low-pressure cylinder, and p 2
and V 2 represent condition of air entering high-pressure
cylinder, and p 3 be final delivery pressure of air, then
the total work done for compression and delivery of air per cycle will be the sum of the work done in each cylinder. Work done in L.P. cylinder is shown by the area pj-1-2’-p2 and in H.P. cylinder by the area p2-2-3-p3 (fig. 11-7). The saving of work done due to imperfect intercooling is shown by the shaded area 2-3-3’-2’. Hence, from eqn. (11.2),
Work required in L.P. cylinder per cycle = —r p iv i
c 1 n
- 1 W
Joule.
Work required in H.P. cylinder per cyde = ——- pzvs.
Total work required per cycle,
' p 3 ' i y f * j
n z _ l n
W =
n n - 1 P ih
n - 1 P P 1
n - 1 w
W x N indicated power of the compressor = — - - — J/sec. or watt.
Joule 0 1 -®a)
where N = No. of cydes per min. = r.p.m. for single-acting compressor, and W = work required in Joules per cycle.
AIR COMPRESSORS 349
If p i v i and P 2 V 2 in eqn. (11.8a) are substituted by mRTi and mRT 2 respectively,
then work done per kg of air can be written as
W = n n - 1
RT\
n - 1 £ v*1,
- 1 + RT
n - 1 n
If the compression is isentropic, y should be substituted for n in eqns. (11.8a) and (11.8b). 11.4.2 Perfect-lntercooling : If intercooling is perfect or complete (fig. 11-6), the point 2 will lie on the isothermal line, i.e., point 2 will coincide with point 2", then P1V1 = P2V2.
Substituting this in eqn. (11.8a), total work required per cycle,
W
n n - 1
pi V 1
n - 1 n - 1 'p z '' lv P V
f -
P 3 v^02 ,
W x N Indicated power of the compressor = — —— J/sec. or W.
where, N = no. of cycles per min. = r.p.m. for single-acting compressor, and W m work required in Joules per cycle. If p i v i in eqn. (11.9a) is substituted by mRTi, then work required per kg of air may be written as
n- 1 n - 1
n — 1 1
'P
w
n
n
- 2 Joules. - <119b> - Indicated power of the compressor = W x mass of air delivered per second J/sec or watt. ... (11.9c) where, W = work required in Joule per kg of air. Referring to fig. 11-6,
Heat rejected to intercooler per min. = mkp (Td - T 2 ) kJ ... (11.10)
where, m = mass of air compressed per minute, kp = specific heat of air at constant pressure, T2’ = temperature of air before entering the intercooler, and T2 = temperature of air after leaving the intercooler. 11.4.3 Ideal Intercooler Pressure : It may be noted from fig. 11-6 thatsaving in work increases as intercooling is increased. When intercooling is perfect, i.e., when air is cooled to intake temperature in the intercooler, point 2 lies on isothermal curve and there is maximum saving. In this case, work required is given by eqn. (11.9a). It may be further noted that this saving in work required also varies with the chosen intercooler
pressure p 2 - When the initial pressure p i and final pressure ps are fixed, the best
value of the intercooler pressure p 2 shall be fixed to give minimum work. This value
If p i v i in eqns. (11.12a) and (11.12b) is substituted by mRTi, then minimum work required per kg of air may be written as
AIR COMPRESSORS 351
n - 1 n ( n W = 2 — — RT< - - -1 Joule n - 1 Pi
...(11.13a)
L \ /
n - 1
and .. (11.13b)
Minimum indicated power of the compressor = W x mass of air delivered per sec. J/sec. or W ..(11.13c) Thus, conditions for maximum efficiency or minimum work required are : .. The air is cooled to initial (intake) temperature in the intercooler, i.e., intercooling is perfect. .. The pressure ratio in each stage is the same. .. The work required for each stage is the same. These conditions can also be extended for three-stage air compressors. 11.5 Three-stage Air Compressor A three-stage air compressor with L.P. and I.P. intercoolers is shown in fig. 11-8. In fig. 11-9, the combined indicator diagram for a three-stage air compressor is shown. The air having volume vi and pressure p i is compressed polytropically to
pressure p 2 in the first or low-pressure cylinder, then delivered through L.P. intercooler
to the second or intermediate pressure cylinder at pressure p 2 , its volume shrinking to
V 2. The air having volume V 2 and pressure p 2 is then compressed polytropically to p$
in the I.P. cylinder, and is then delivered through I.P. intercooler to the high-pressure
cylinder (third cylinder) at pressure p 3 , its volume shrinking to vs. This air having
pressure pa and volume v 3 is then compressed polytropically to volume V 4 in the H.P.
cylinder and then delivered to the receiver at pressure p 4 -
As in the case of two-stage compression, the shaded area, in fig. 11-9 represents the saving of work due to using three cylinders with inter-stage cooling instead of single-stage. (a) Work required per cycle when intercooling is imperfect, i.e., air is not cooled to intake temperature in the intercoolers,
If pivi, P 2 V 2 , and P 3 V 3 in eqn. (11.14a) are substituted by mRTi, mRT 2 and mRT 3
respectively, then work required per kg of air can be written as
n - 1 + ^ - P3V* Joule .. (11.14a)
(^352) ELEMENTS OF HEAT ENGINES Vol.l
W =
n n -
n- 1 P? L V
n n -1 RTz
n - 1 P
n n -
RT 3
n - 1 P4 (^) -
Joule .. (11.14b)
c
PHtvtrytof«c»<v»f tmmu* leuflH
® < S s )
- •^ JM.R^ pitlon^ J thro* Hr
A«uct>o (^) n^ \T vOlvf.M il..^1 I *
I
f p f e n
i
« f
LP.
piston
■ucfton Volvo
I J
t
fWolff
•> -. ■8. 9 e. •. it:'- st o;. t. i t
t i.
V tfn
i f 3 A!
OfOtn cocks
Fig. 11-8. Arrangement of three-stage air compressor with intercoolers.
to t* Dull*
Volwmo Fig 11-9. Indicator diagram of three-stage compression with perfect intercooling.
(b) When intercooling is perfect, ptV i = P2V 2 - P3V 3 Hfe 11-9).
Substituting pivi lor p2v2 and P 3 V 3 in eqn. (11.14a), work required per cycle,
W
n - 1 n - 1 n -^^1 n p i _v_ n - 1
fpz n + _(f )_* n + n - 3 Joule [ p i J (^) W w _ ... (11.15a) If p/Vf in eqn. (11.15a) is substituted by mRTh then work required per kg air may be written as,
n - 1 n -^1 n - 1 = n~. RT
n - 1
' Pz' [ p i j
n
n
n
- 3 Joule^ .. (11.15b)
(c) Work required is minimum when Pz _ 03 _ Pa Pi “ Pz ~ P
All the expression derived above for work required refer to the work actually done or required to be done on the air, and the power derived from these expressions will be referred to as indicated power or air power. 11.5.1 Advantages of multi-stage compression : The advantages of multi-stage compression are as follows :
- Reduction in power required to drive the compressor owing to compression being approximated to isothermal,
- Better mechanical balance of the whole unit and uniform torque,
- Increased volumetric efficiency as a result of the lower delivery pressure in the L.P. cylinder clearance space,
- Reduced leakage loss owing to reduced pressure difference on either sides of the piston and valves,
- Less difficulty in lubrication due to the lower working temperature, and
- Lighter cylinders. 11.6 A ir Compressor Terminology The following terminology should be well understood before attempting to estimate the performance of the air compressor. Free air delivered is the volume of air delivered under the conditions of temperature and pressure existing at the compressor intake, i.e., volume of air delivered at surrounding air temperature and pressure. In the absence of any given free air conditions, these are generally taken as 1 01325 bar and 15‘C. Capacity of a compressor is the quantity of the free air actually delivered by a compressor in cubic metres per minute. Piston displacement is the volume in cubic metre (m ) obtained asthe product of the piston area in m2 and the piston stroke in metre. Displacement per minute is the product of the piston displacement and working strokes per minute. For multi-stage compressors, the displacement is based on low-pressure cylinder only, since it determines the amount of air passing through the other cylinder. Indicated power or air power is the power determined from the actual indicator diagram taken during a test on the compressor. It is calculated in the same manner as is done in the case of a steam engine and internal combustion engine. Shaft or brake power is the power delivered to the shaft of the compressoror the power required to drive the compressor. The compressor may be driven by an engine or an electric motor. [Shaft or brake power] - [Air or indicated power] = [Friction power] , „... „.. Air (indicated) power and Mechanicalefficiency, rim = o. -. —. - , -------- 1 1 Shaft (brake) power Isothermalpower of acompressor is calculated from the theoretical indicator diagram drawn on the basis of an assumption that the compression is isothermal. (a) Referring to eqn. (11.7a) for a single-stage compressor without clearance,
354 ELEMENTS OF HEAT ENGINES Vol.l
Isothermal work required per cycle, _W = p_ vi loge '( * Pi Jou,e^ .. (11.19a)
Isothermal power = _pw _ log©
AIR COMPRESSORS
v * ,
x J/sec. or W 60
.. (11.19b)
where N = no. of cycles per minute. If p i v i in eqn. (11.19a) is substituted by mRTi, then isothermal work required per kg of air may be written as,
W = RTy log, Joule
Isothermal power » W x mass of air delivered per sec. J/sec. or W (b) For a two-stage compressor,
Isothermal work required per cycle, W = pi vi loge
N Isothermal power = _pi v_ loge v vP1,
V * ,
Joule
x - - J/sec. or W 60
(11.20a)
(11.20b)
(11.21a)
(11.21b)
If p i v i in eqn. (11.21a) is substituted by mRTi, then isothermal work required per kg of air may be written as .. (11.22a) W = RTi \oge w V * /
Joule
Isothermal power = W x mass of air delivered per sec. J/sec. or W .. (11.22b) (c) Similarly, for a three-stage compressor, .. (11.23a) Isothermal work required per cycle, W = pi vi log
N Isothermal power = pi _v_ ioge
*/ _ ,* P
fp ^
S ',
Joule
,P1v
x — J/sec. or W
.. (11.23b)
where N * no. of cycles per min. Adiabatic power is calculated from a theoretical indicator diagram drawn on the basis of an assumption that the compression is an ideal adiabatic, i.e., isentropic.
Adiabatic work required, W = (pzvz-pivtf Joule per cyde
This equation for the adiabatic work required may be expressed in more convenient form by writing its equivalent,
W = mR(Tz - T O = mRT,
f Tr ' f r '
Tz Since -=- = » (^1) v * /
I z l
Adiabatic work required, W fPz
i -_ 1 W = —Y— mRT} y -
W
Joule per cycle
AIR COMPRESSORS 357
- Very high speed of rotation. With the decrease of volumetric efficiency, the capacity (quantity of free air delivered) of the compressor decreases.
Problem-1 : A single-cylinder, single-acting reciprocating air compressor has a cylinder of 24 cm diameter and linear piston speed of 100 metres per minute. It takes in air at 100 kPa (100 kN/m2) and delivers at 1 MPa ( 1 MN/rrP), Determine the indicated power of the compressor. Assume the law of compression to be pv ture of air at inlet is 288 K. Neglect clearance effect.
1. ■ constant. The tempera-
Given p i = 100 kPa « 100 x 10 Pa; p2 = 1 MPa = 1,000 kPa - 1,000 x 103 Pa.
P* Pi
1,000 x 10J 1, 100 x 10J 100
Swept volume in m 3 /min. - - x I x r.p.m.
(where, I - piston strokes in metre, and d= diameter of the cylinder in metre) ** It Swept volume = ^ x f 24 ' 100 /
100 3.. x m /min.
( v piston speed = 2 x I x r.p.m. = 100 metres/min.) _... 3. 2-261 3. = 2-261 m /min. = m /sec.
Referring to fig. 11-10 and using eqn. (11.2), Work required per sec.,
W = pyV] x
n n - 1
n - 1
L V^ S ' /
- 1 (^) J/sec.
[where, p i is pressure in Parcals (Pa) and volume of air compressed, vi is in m per sec.]
W = (too x to3) x 2f l X ' ±^000 )
100 \ /
1 2 5 - 1 25
= 1,88616 x (1 5848 - 1) - 11,030 J/sec. or 11,030 W .-. Indicated power of the compressor = 11,030 W i.e. 11-03 kW Problem-2- : A single-acting, single-stage air compressor developing indicated power of 11 kW, runs at 200 r.p.m. and has a linear piston speed of 100 metres per min. If the suction pressure and temperature are 100 kPa and 15‘C respectively and delivery pressure is 1,000 kPa, calculate the dimensions of the compressor cylinder. Assume the law of compression to be pv125 = constant. Neglect clearance effects. Referring to fig. 11-10, and considering polytropic compression 1-2, p iv in = p 2 V2n,
358 ELEMENTS OF HEAT ENGINES Vol.l
V 1 _ n^ 10' Vz (^) [p .j (^) V 1 /
1* = 6 31
P2VZ - P1V
Using eqn. (11.1a), work done per cycle, W = pzvz + - — ------- p iv i
^ p Work done per cycle in kJ, W ^pa Displacement volume in m ,-3 vi
P2 vz
p z v z - p i vi pzva+ p in
V 1 V^1
-P1 +
p z y z Vi - P i
1
1,000 x 1
= 1,000 x £ “ -1 0 0 +
Piston stroke, I = piston speed per min.
n - 1
= 292 4 kPa
100 piston strokes per min. (2 x r.p.m.) 2 x 200 Indicated power of compressor = pm x I x a x n watt. 200N
= 0-25 metre or 25 cm
where n = no. of cycles per sec. ^ 60
i.e. 11 x 103 - 292 4 x 103 x 0 25 x 0-7854 x
11 x 103 x 104 x 60
r < n 100
292 4 x 103 x 0-25 x 0-7854 x 200
d = V574 79 = 23-98 cm. Problem-3 : A single-acting, single-stage air compressor is belt driven from an electric motor at 300 r.p.m. The cylinder diameter is 20 cm and the stroke is 24 cm. The air is compressed from one atmosphere to 8 atmospheres and the law of compression is pv125 = constant. Find the power of the electric motor if the transmission efficiency is 96 per cent and the mechanical efficiency of the compressor is 85 per cent Neglect clearance effect. Swept volume,
f 20 100
\
X
= 000754 m
Referring to fig. 11-11 and using eqn. (11.2), work required per cycle,
... n W = pi vi n - 1
n - 1 P P
- 1 Joule.
x (101325 x 105) x 0 00754 x
1 -2 5 - 1 25
- 1
= 5 x (1 01325 x 105) x 0-00754 x (1-515 - 1) - 1,967-27 Joule/cycle.
Given: p i = 1 bar = 1 x 10s Pa; p2 - p2 = 7 bar = 7 x 10s Pa; — = -= = __ Pi 1 x 1 0 5 1 (a) Isentropic compression : '
Referring to fig. 11-12 and using eqn. (11.6a), isentropic work required per sec.,
360 ELEMENTS OF HEAT ENGINES Vol.l
W = pi vi Y~ *
I r l Y
LVP1V
-1 J/sec.
—— — x (1 x 1 -4 _ 1 u (^) x 10®) x lu } (^) x — 60 —
'7 X
v /
14
■ x <1 x 105) x x [1 7 4 4 -1 ]
Fig. 11 12. Single stage compression. (^) = 73,750 J/sec. or 73,750 W
/. Power required to drive the compressor = 73,750 W i.e. 73 75 kW No heat is rejected during isentropic compression. (b) Isothermal Compression Referring to fig. 11-12 ?nd using eqn. (11.7a), isothermal work required/sec., fPz P^
W = p i V) log< = (1 x 105) x ~ x loge
= (1 x 105) x x 1-9459 = 55,080 J/sec. or 55,080 W 60 Required power input = 55,080 W i.e. 55-08 kW In isothermal compression, as the temperature remains constant, there is no change in internal energy and the entire work of compression, i.e. 55-08 kJ/sec. is rejected to jacket cooling water. Heat rejected during isothermal compression = 55-08 kJ/sec. Problem-6 : A two-stage air compressor delivers 145 m3 of free air per hour. The pressure and temperature in the cylinder at the start of compression are 1 bar and 34'C respectively. The diameter of the low-pressure cylinder is twice that of the high pressure cylinder. The air enters the high-pressure cylinder at a temperature of 40’C and is then compressed to 17-5 bar, the law of compression being pv1 « constant for both stages. Neglecting the effects of clearance, estimate :■ (a) the intercooler pressure, (b) the air power required, and (c) the ratio of cylinder diameters for minimum work making the usual assumptions regarding the intercooler conditions. The free air conditions are 101325 bar and 15'C. Take R = 0287 kJ/kgK for air. (a) Since, the diameter of L.P. cylinder is twice that of the H.P. cylinder, the ratio of L.P. to H.P. cylinder volumes will be 4, i.e. vt = 4v^ Applying characteristic equations at points of suction in L.P. and H.P. cylinders,
P 1 V 1 P2V
AIR COMPRESSORS
i.e.
1 x 4 v 2 P 2 x vz h T
Intercooler pressure, pz =
1 x 4 x 313 307 = 4 078 bar
(b) Using characteristic equation for free air conditions, pv - mRT
i.e. (1 01325 x 10s) x —J f = m x (0287 x 103) x (15 + 273) 60 m = 2 963 kg of air per minute. Referring to fig. 11-13, and using eqn. (11.8b) for imperfect intercooling,
....... ... fWork required in] , [Work required ini Total w o * required per kg of air, W = | Hp ^ |jnder j +j LP J
.-. W = RTi - — n — 1
c t
1
|c
[ p i J
n n - 1
n - 1 P3^ n
v /
= R
n n - 1
n - 1 n - 1
V
"P2^
W
n
- (^1) + r2 M W
n
- 1
= 0-287 x 307 1 22 1 2 2 - 1 = 290 kJ per kg of air
v
0-
1-
0 22 f I 7.5 \ l - 4078
o.ggo Work done per sec. = W x m = 290 x ——- = 14-32 kJ/sec. 60 Air power = 14-32 kW
Fig. 11-13. Two-stage compression with imperfect intercooling
Fig. 11-14.
